Explanation Given information: The equation of line is x 3 = y − 1 2 = z 1 and point is P ( 1 , 0 , 1 ) Formula Used: Distance of a point ( x 0 , y 0 , z 0 ) from the line x − x 1 a = y − y 1 b = z − z 1 c is d = ‖ v × Q P ‖ ‖ v ‖ . Calculation: It is given that equation of line is x 3 = y − 1 2 = z 1 The point is P ( 1 , 0 , 1 ) = ( x 0 , y 0 , z 0 ) Distance of a point from line is given by d = ‖ v × Q P ‖ ‖ v ‖ On comparing the equation line x 3 = y − 1 2 = z 1 with x − x 1 a = y − y 1 b = z − z 1 c a = 3 , b = 2 & c = 1 and line contain the point Q ( x 1 , y 1 , z 1 ) = ( 0 , 1 , 0 ) Therefore Q P = ( 0 − 1 ) i + ( 1 − 0 ) j + ( 0 − 1 ) k Q P = − i + j − k & v is the vector parallel to given linegiven by v = 3 i + 2 j + k Distance between point P ( 1 , 0 , 1 ) = ( x 0 , y 0 , z 0 ) and line

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 9.6, Problem 30PS

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To find: Distance of a point P(1,0,1) from a line x3=y−12=z1 .

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Chapter 9.6, Problem 29PS

Chapter 9.6, Problem 31PS

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Chapter 9 Solutions

Calculus

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Distance of a point P ( 1 , 0 , 1 ) from a line x 3 = y − 1 2 = z 1 .

Solution Summary: The author calculates the distance of a point P(1,0,1) from the equation x3.

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To find: Distance of a point P(1,0,1) from a line x3=y−12=z1 .

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Chapter 9 Solutions